Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

Question

Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

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In Algebra by Artin, he says this proposition is proven analogously to:

Here, he shows that if $n$ is prime, then $\mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.

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The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.

Answer 1

For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.

Now let

$M \subset F[x] \tag 1$

be a maximal ideal; since $F[x]$ is a principal ideal domain, we have

$M = (m(x)) \tag 2$

for some

$m(x) \in F[x]; \tag 3$

we may clearly take $m(x)$ to be monic, since the leading coefficient $\mu$ of $m(x)$, satisfying as it does $\mu \ne 0$, is a unit; thus $\mu^{-1} m(x)$ is monic and

$(\mu^{-1} m(x)) = (m(x)); \tag 4$

now if $m(x)$ were reducible in $F[x]$, we would have

$m(x) = p(x)q(x), \; p(x), q(x) \in F[x], \; \deg p(x), \deg q(x) \ge 1; \tag 5$

consider the ideal

$(p(x)) \subsetneq F[x]; \tag 6$

it is clearly proper: since $\deg p(x) \ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$ other than $0$. Also,

$(m(x)) = (p(x)q(x)) \subsetneq (p(x)), \tag 7$

for

$p(x) \notin (p(x)q(x)) \tag 8$

lest for some

$r(x) \in F[x] \tag 9$

we have

$p(x) = r(x)p(x)q(x), \tag{10}$

or

$p(x)(r(x)q(x) - 1) = 0, \tag{11}$

whence

$r(x)q(x) = 1, \tag{12}$

which yields

$\deg r(x) + \deg q(x) = \deg 1 = 0, \tag{13}$

impossible in light of the assumption $\deg q(x) \ge 1$; thus we have shown that

$(m(x)) = (p(x)q(x)) \subsetneq (p(x)) \subsetneq F[x] \tag{14}$

in the event that $m(x)$ is reducible, which further shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.

Answer 2

There's a classic result in commutative algebra that you can apply:

Let $B$ be an integral domain, $A$ be a subring such that $B$ is integral over $A$. Then $B$ is a field if and only if $A$ is a field.

Answer 3

In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.

If reducible, then we have a zero divisor, so it can't be a field. If irreducible, then all polynomial  can be subjected to Euclidean algorithm which gives you a multiplicative inverse.

The remaining field axioms follow from the fact that we have a ring quotient.

Answer 4

It is really easy to show that $\mathbb{Z}/p\mathbb{Z}$ is a field when $p$ is prime, because it easily stems from Bézout's identity.

However, Bézout's identity also holds over any Euclidean domain. Let $g(x)\in F[x]$, with $g(x)$ not divisible by $f(x)$. We want to show that $g(x)+(f)$ is invertible in $F[x]/(f)$.

Since $f$ is irreducible, $1$ is a greatest common divisor of $f$ and $g$, hence there are polynomials $u,v\in F[x]$ such that $$ 1=f(x)u(x)+g(x)v(x) $$ Then, clearly, $v(x)+(f)$ is the inverse of $g(x)+(f)$ in $F[x]/(f)$.

Note that we actually proved that, for every irreducible element $n$ in a Euclidean domain $R$, the quotient ring $R/(n)$ is a field.